Question

Ok I have a form with multiple submit buttons.
The coding on my php file has a header with a url depending on which form was entered. My issue is when I submit the form( no matter which button I use) the window that pops up is not the url action assigned to that button but the php file itself. What am I doing wrong?

the form starts of like this so that you can see if I directed it correctly

<form method="post" action="http://gamerzacademy.com/foodCYO.php" target="_blank">
<input type="text" name="uid">
<input type="submit" name="Dish1" value="Dish1" onclick="
  this.disabled=true; 
  this.value='Gift Opened'; 
  document.FreeFoodForm.submit();">
<input type="submit" name="Dish2" value="Dish2" onclick="
  this.disabled=true; 
  this.value='Gift Opened'; 
  document.FreeFoodForm.submit();">

etc……

now the php file starts like this

<?php
if ($_REQUEST['Dish1'] == "Dish1") {
    header("Location: url1".urlencode($_POST['uid']));
}
else if ($_REQUEST['Dish2'] == "Dish2") {
    header("Location: url2".urlencode($_POST['uid']));
}
else if ($_REQUEST['Dish3'] == "Dish3") {
    header("Location: url3".urlencode($_POST['uid']));
}
.....etc
?>

Answer 1

You are posting the form through Javascript. The code doesn’t know which button was clicked, so the value of that button isn’t posted to the form. Therefor, your form cannot see which button was clicked. If you change the method to get, you will see which value do or do not get posted.

I think you don’t need to post from Javascript at all. Just let the button do the posting. Only the name and value of the button that was clicked will be posted.

B.t.w., you disable the button, presumably because you don’t want people to press the button twice, but in your setup they still can press any other button. I think it is wise to disable all of them.